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  • srand[getpid[]] /* initialize some of the memory */ memset[heightmap, 0, MAPSIZE*MAPSIZE]

    srand[getpid[]] /* initialize some of the memory */ memset[heightmap, 0, MAPSIZE*MAPSIZE] memset[vpage, 0, RENDERWIDTH * RENDERHEIGHT] printf["Creating dx d fractal terrain\n", MAPSIZE, MAPSIZE] heightmap[0] = [rand[] 128] + 64 // initialize starting point on map CreateFractalMap[0, 0, MAPSIZE, MAPSIZE] printf["Smoothing terrain\n"] for [i = 0 i < 5 i++] SmoothMap[] MakeColorMap[]

    標(biāo)簽: MAPSIZE initialize heightmap getpid

    上傳時(shí)間: 2013-12-25

    上傳用戶:CSUSheep

  • 找一個(gè)最小的自然數(shù)

    找一個(gè)最小的自然數(shù),使它等于不同的兩組三個(gè)自然數(shù)的三次冪之和,即找最小的x,使得:x=a*a*a+b*b*b+c*c*c = d*d*d+e*e*e+f*f*f 其中,a,b,c,d,e,f都是自然數(shù),a<=b<=c, d<=e<=f [a,b,c]!=[d,e,f] 進(jìn)一步,是否還存在另外一個(gè)自然數(shù)滿足上述條件,可能的話請輸出其結(jié)果

    標(biāo)簽:

    上傳時(shí)間: 2017-05-16

    上傳用戶:vodssv

  • P3.20. Consider an analog signal xa (t) = sin (2πt), 0 ≤t≤ 1. It is sampled at Ts = 0.01, 0.05, and

    P3.20. Consider an analog signal xa (t) = sin (2πt), 0 ≤t≤ 1. It is sampled at Ts = 0.01, 0.05, and 0.1 sec intervals to obtain x(n). b) Reconstruct the analog signal ya (t) from the samples x(n) using the sinc interpolation (use ∆ t = 0.001) and determine the frequency in ya (t) from your plot. (Ignore the end effects.) C) Reconstruct the analog signal ya (t) from the samples x (n) using the cubic spline interpolation and determine the frequency in ya (t) from your plot. (Ignore the end effects.)

    標(biāo)簽: Consider sampled analog signal

    上傳時(shí)間: 2017-07-12

    上傳用戶:咔樂塢

  • g a w k或GNU awk是由Alfred V. A h o

    g a w k或GNU awk是由Alfred V. A h o,Peter J.We i n b e rg e r和Brian W. K e r n i g h a n于1 9 7 7年為U N I X創(chuàng)建的a w k編程語言的較新版本之一。a w k出自創(chuàng)建者姓的首字母。a w k語言(在其所有的版本中)是一種具有很強(qiáng)能力的模式匹配和過程語言。a w k獲取一個(gè)文件(或多個(gè)文件)來查找匹配特定模式的記錄。當(dāng)查到匹配后,即執(zhí)行所指定的動(dòng)作。作為一個(gè)程序員,你不必操心通過文件打開、循環(huán)讀每個(gè)記錄,控制文件的結(jié)束,或執(zhí)行完后關(guān)閉文件。

    標(biāo)簽: V. Alfred GNU awk

    上傳時(shí)間: 2014-01-02

    上傳用戶:hwl453472107

  • RSA ( Rivest Shamir Adleman )is crypthograph system that used to give a secret information and digit

    RSA ( Rivest Shamir Adleman )is crypthograph system that used to give a secret information and digital signature . Its security based on Integer Factorization Problem (IFP). RSA uses an asymetric key. RSA was created by Rivest, Shamir, and Adleman in 1977. Every user have a pair of key, public key and private key. Public key (e) . You may choose any number for e with these requirements, 1< e <Æ (n), where Æ (n)= (p-1) (q-1) ( p and q are first-rate), gcd (e,Æ (n))=1 (gcd= greatest common divisor). Private key (d). d=(1/e) mod(Æ (n)) Encyption (C) . C=Mª mod(n), a = e (public key), n=pq Descryption (D) . D=C° mod(n), o = d (private key

    標(biāo)簽: crypthograph information Adleman Rivest

    上傳時(shí)間: 2017-09-01

    上傳用戶:chfanjiang

  • fortan程序數(shù)值分析之Newton法求解非線性方程組

    module M_GAUSS !高斯列主元消去法模塊 contains subroutine LINEQ(A,B,X,N) !高斯列主元消去法    implicit real*8(A-Z)    integer::I,K,N    integer::ID_MAX  !主元素標(biāo)號    real*8::A(N,N),B(N),X(N)    real*8::AUP(N,N),BUP(N)    !A,B為增廣矩陣    real*8::AB(N,N+1)    real*8::VTEMP1(N+1),VTEMP2(N+1)    AB(1:N,1:N)=A    AB(:,N+1)=B

    標(biāo)簽: fortan Newton 程序 數(shù)值分析 方程 非線性

    上傳時(shí)間: 2018-06-15

    上傳用戶:answer123

  • 數(shù)組子系統(tǒng)

    #include <stdio.h> #include <stdlib.h> #define SMAX 100 typedef struct SPNode { int i,j,v; }SPNode; struct sparmatrix { int rows,cols,terms; SPNode data [SMAX]; }; sparmatrix CreateSparmatrix() { sparmatrix A; printf("\n\t\t請輸入稀疏矩陣的行數(shù),列數(shù)和非零元素個(gè)數(shù)(用逗號隔開):"); scanf("%d,%d,%d",&A.cols,&A.terms); for(int n=0;n<=A.terms-1;n++) { printf("\n\t\t輸入非零元素值(格式:行號,列號,值):"); scanf("%d,%d,%d",&A.data[n].i,&A.data[n].j,&A.data[n].v); } return A; } void ShowSparmatrix(sparmatrix A) { int k; printf("\n\t\t"); for(int x=0;x<=A.rows-1;x++) { for(int y=0;y<=A.cols-1;y++) { k=0; for(int n=0;n<=A.terms-1;n++) { if((A.data[n].i-1==x)&&(A.data[n].j-1==y)) { printf("%8d",A.data[n].v); k=1; } } if(k==0) printf("%8d",k); } printf("\n\t\t"); } } void sumsparmatrix(sparmatrix A) { SPNode *p; p=(SPNode*)malloc(sizeof(SPNode)); p->v=0; int k; k=0; printf("\n\t\t"); for(int x=0;x<=A.rows-1;x++) { for(int y=0;y<=A.cols-1;y++) { for(int n=0;n<=A.terms;n++) { if((A.data[n].i==x)&&(A.data[n].j==y)&&(x==y)) { p->v=p->v+A.data[n].v; k=1; } } } printf("\n\t\t"); } if(k==1) printf("\n\t\t對角線元素的和::%d\n",p->v); else printf("\n\t\t對角線元素的和為::0"); } int main() { int ch=1,choice; struct sparmatrix A; A.terms=0; while(ch) { printf("\n"); printf("\n\t\t      稀疏矩陣的三元組系統(tǒng)       "); printf("\n\t\t*********************************"); printf("\n\t\t      1------------創(chuàng)建          "); printf("\n\t\t      2------------顯示          "); printf("\n\t\t      3------------求對角線元素和"); printf("\n\t\t      4------------返回          "); printf("\n\t\t*********************************"); printf("\n\t\t請選擇菜單號(0-3):"); scanf("%d",&choice); switch(choice) { case 1: A=CreateSparmatrix(); break; case 2: ShowSparmatrix(A); break; case 3: SumSparmatrix(A); break; default: system("cls"); printf("\n\t\t輸入錯(cuò)誤!請重新輸入!\n"); break; } if (choice==1||choice==2||choice==3) { printf("\n\t\t"); system("pause"); system("cls"); } else system("cls"); } }

    標(biāo)簽: 數(shù)組 子系統(tǒng)

    上傳時(shí)間: 2020-06-11

    上傳用戶:ccccy

  • 大尺寸液晶電視用LED背光源的設(shè)計(jì)與制作.pdf

    摘# 要:設(shè)計(jì)和制作了一款&& ?G(!& )*)液晶電視用4F9 背光源。模擬出4F9 的光學(xué)分布,以此為基礎(chǔ)模擬出4F9 陣列的光強(qiáng)和顏色分布,得到適合的背光源厚度尺寸。在實(shí)際制作中,采用高效的驅(qū)動(dòng)電路對4F9 陣列進(jìn)行驅(qū)動(dòng),利用鋁制散熱片為背光源提供必須的散熱。測試的結(jié)果,在整體背光源功耗為"$% M 時(shí),中心亮度達(dá)到"D DE% ?6 N G!,均勻度為CO@ " P,色彩還原性達(dá)到=QR’ 標(biāo)準(zhǔn)"%! P,遠(yuǎn)遠(yuǎn)超過’’S4 背光源的A% P。

    標(biāo)簽: led 光源

    上傳時(shí)間: 2021-12-09

    上傳用戶:

  • dsp hư ớ ng dẫ n giao tiế p LCD code viế t bằ ng C

    dsp hư ớ ng dẫ n giao tiế p LCD code viế t bằ ng C

    標(biāo)簽: 7871 7899 7851 7857

    上傳時(shí)間: 2017-06-17

    上傳用戶:xuanjie

  • 四柱漢諾塔問題的求解程序.解題思路:如a,b,c,d四柱. 要把a(bǔ)柱第n個(gè)盤移到目標(biāo)柱子(d柱),先把上層 分兩為兩部份,上半部份移到b柱,下半部分移到c柱,再把第n盤移到 目標(biāo)柱子,然后,c柱盤

    四柱漢諾塔問題的求解程序.解題思路:如a,b,c,d四柱. 要把a(bǔ)柱第n個(gè)盤移到目標(biāo)柱子(d柱),先把上層 分兩為兩部份,上半部份移到b柱,下半部分移到c柱,再把第n盤移到 目標(biāo)柱子,然后,c柱盤子再移到目標(biāo)柱子,再把b柱盤子移到目標(biāo)柱子. 細(xì)節(jié)地方: 上半部份移到b柱時(shí),它的中間變量柱子是有二選一的.而下半部分 移到c柱時(shí),它的中間變量柱子只有一個(gè)(因?yàn)橐粋€(gè)柱子已被上半部份 占了).b,c也移到目標(biāo)柱子時(shí)同理。

    標(biāo)簽: 程序

    上傳時(shí)間: 2013-12-22

    上傳用戶:aeiouetla

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